Multiplying A Coefficient by an Indexed Multiplier using Generating Functions

Question

If I have a particular exponential generating function,

$$G(x)=\sum_{n=0}^\infty a_n\frac{x^n}{n!}$$

then what would be the generating function for

$$H(x)=\sum_{n=0}^\infty (n+1)a_n\frac{x^n}{n!}$$ in terms of $G(x)$? I know that I can change it to

$$H(x)=\sum_{n=0}^\infty na_n\frac{x^n}{n!}+\sum_{n=0}^\infty a_n\frac{x^n}{n!}=J(x)+G(x)$$

Now I know that $$G'(x)=\sum_{n=0}^\infty na_n\frac{x^{n-1}}{n!}$$ and so

$$H(x)=xG'(x)+G(x)$$

Is there a way to write $H$ without the derivative of $G$?

I would like $H$ as a function in terms of strictly $G$, and without a $G'$ term.

Answer 1

This will not be possible in general. Indeed, if you have both $H(x) = \Phi(x,G(x))$ and $H(x) = xG'(x) + G(x)$ for all $x$ (where $\Phi$ is a nice, regular function), then you can derive an expression of the form $\Phi(x, G(x)) = xG'(x) + G(x)$ for all $x$, i.e. a differential equation of the form $$G^\prime(x) = \Psi(x,G(x)).$$ (where $\Psi$ is regular/"smooth enough"). Along with $G(0)$ and $G^\prime(0)$ (i.e., only $a_0$ and $a_1$), this differential equation will then entirely characterize $G$ -- while you are hoping to get a general formula that works for any $G$ defined by its generating sequence $(a_n)_n$.

Answer 2

It shouldn't be possible to give a more simplified expression which is valid in general, since both $G(x)$ and $G'(x)$ depend on the choice of coefficients $(a_n)_n$ (albeit in a different manner).

For example, let $G_1(x)=e^x$ and $G_2(x)=\sin x$. Then:

$$H_1(x)=(x+1)G_1(x)$$

$$H_2(x)=xG_2(\frac{\pi}{2}-x)+G_2(x)$$

Thus while knowledge of $(a_n)_n$ should allow you in many cases to write $H(x)$ solely in terms of $G(x)$, how this will be accomplished will depend specifically on the choice of $(a_n)_n$, and as the example above shows, the results from different choices will not necessarily be reconcilable with each other, i.e. since the above two formulas (I believe) cannot be writted as special cases of a formula involving only $G(x)$ and not $G'(x)$.